Question

A spring of spring constant $5\times {10}^{3}N/m$ is stretched initially by $5cm$ from the unstretched position. Then the work required to stretch it further by another $5cm$ is-

• A. $18.75N-m$
• B. $25.00N-m$
• C. $6.25N-m$
• D. $12.50N-m$

Answer 1

Answer: (C)

$6.25N-m$

Spring constant $k=5\times {10}^{3}N/m$

Displacement $x=10cm=10\times {10}^{-2}m={10}^{-1}m$

Work required$=PE=\frac{1}{2}k(x-x_0{)}^2$ where $x_0$ is $x_0=5cm=5\times {10}^{-2}m$

$PE=\frac{1}{2}\times 5\times {10}^3\times 25\times {10}^{-4}$

$=62.5\times {10}^{-1}$

$=6.25J$ or $6.25N-m$